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hodyreva [135]
3 years ago
13

Consider a horse pulling a buggy. Is the

Physics
1 answer:
Degger [83]3 years ago
8 0

Yes, it's true.

But 2nd Newton Law always come to play when the horse is to move forward because obviously the forces interact antagonistically and mass has to be accounted for.

That's what I think. Hope it's right, all the best.

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A 68.2 kg base runner begins his slide into second base while moving at a speed of 4.33 m/s. He slides so that his speed is zero
qwelly [4]

Answer:

147.653J = W_f

Explanation:

We know that:

E_i - E_f = W_f

where E_i is the initial energy, E_f is the final energy and W_f is the energy lost due to friction.

so:

\frac{1}{2}MV^2-0 = W_f

Where M is the mass and V is the velocity of the runner, so:

\frac{1}{2}(68.2kg)(4.33m/s)^2 = W_f

147.653J = W_f

7 0
4 years ago
Differentiate between loadstone and bar magnet​
MakcuM [25]
As nouns the difference between magnet and lodestone




is that magnet is a piece of material that attracts some metals by magnetism while lodestone is a naturally occurring magnet.
7 0
2 years ago
Aranza is 17 years old. What is her Maximum Heart Rate? I will give brainliest ​
tensa zangetsu [6.8K]

Answer:

203 beats per minute

Explanation:

4 0
3 years ago
Read 2 more answers
A box rests on the back of a truck. The coefficient of friction between the box and the surface is 0.32. (3 marks) a. When the t
cricket20 [7]

Answer:

Part a)

here friction force will accelerate the box in forward direction

Part b)

a = 3.14 m/s/s

Explanation:

Part a)

When truck accelerates forward direction then the box placed on the truck will also move with the truck in same direction

Here if they both moves together with same acceleration then the force on the box is due to friction force between the box and the surface of the truck

So here friction force will accelerate the box in forward direction

Part b)

The maximum value of friction force on the box is known as limiting friction

it is given by the formula

F = \mu mg

so we have

F = ma = \mu mg

now the acceleration is given as

a = \mu g

a = (0.32)(9.8) = 3.14 m/s^2

5 0
3 years ago
A ball starts at rest and rolls down a ramp, developing a speed of 45 m/s in 7 seconds. Calculate the ball's acceleration.
Alexxx [7]

Answer:

6.429 m/s^2.

Explanation:

Using equations of motion,

i. vf = vi + at

ii. vf^2 = vi^2 + 2a*S

iii. S = vi*t + 1/2 * (a*t^2)

Where,

vf = final velocity of the motion

vi = initial velocity of the motion

S = distance travelled

t = time taken to complete the motion

a = acceleration due to gravity

Given:

vi = 0m/s

vf = 45 m/s

t = 7 s

a = ?

Using the i. equation of motion,

vf = vi + at

45 = 0 + a*7

a = 45/7

= 6.429 m/s^2

6 0
3 years ago
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